J.V. Matijasevic. Enumerable sets are diophantine. Soviet Math. Dokl., 11:354--358, 1970.  Home/Search   Document Not in Database   Summary   Related Articles   Check

....M to accept) is undecidable, even when restricted to k = 2. 2) The emptiness problem is decidable when k = 1. 3) The emptiness problem is decidable for any k, provided there is only one parameterized constant. Proof: The proof of part 1 uses the undecidability of Hilbert s Tenth Problem (HTP) [Mat70], which is to decide for a given polynomial P (x 1 ; x n ) with integer coefficients whether it has a nonnegative integral root. First consider a term st(x 1 ; x n ) sx n of the polynomial P (x 1 ; x n ) where s = or Gamma, i 1 ; i n 0. We show how to construct a ....

Y. Matijasevic, Enumerable sets are Diophantine, Soviet Math. Dokl, Vol. 11, 354--357, 1970.

....functions in , and relations in , #, # are used. The satisfiability problem deals with the existence of values for free variables which satisfy a given formula. The satisfiability problem is in general undecidable if one considers polynomial formulae on integer variables (see [74]) Di#erent classes are considered to avoid such di#culty. To prove decidability, usually an algorithm based on the quantifier elimination approach is provided. Given a formula with a quantified variable, the quantifier elimination deals with the problem of finding an equivalent formula without ....

....a with the array b. For # #( # id) the satisfiability problem consists in finding whether # is not equivalent to false. Satisfiability of quantified formulae is in general undecidable. This negative result comes from the fact that it is undecidable for integer variables. As an example, in [74] it is proved that checking the emptiness of the set V ( # k ) # = 0) # k ) is undecidable. Therefore the following corollary holds. #( # id) it is undecidable whether # is satisfiable. Satisfiability of quantified formulae on real identifiers (i.e real variables and ....

Matijasevic, J. V.: Enumerable sets are Diophantine. Soviet Mathematics 11, (1970), 354--357.

....two stack cells. This restricted worktape is called a checking tape. Call this machine CCM. CCM acceptors without counters have been studied in [28] Theorem 3 The emptiness problem for CCM acceptors is undecidable. Proof. The proof uses the undecidability of Hilbert s Tenth Problem (HTP) [29], which is to decide for a given polynomial p(x 1 ; xn ) with integer coefficients whether it has a nonnegative integral root. We omit the construction, but the idea is to show that we can effectively construct, given a polynomial p(x 1 ; xn ) a CCM acceptor M p such that p has no ....

Y. Matijasevic, "Enumerable sets are Diophantine," Soviet Math. Dokl, 11 (1970) 354--357.

....accept) is undecidable, even when restricted to k = 2. 2) The emptiness problem is decidable when k = 1. 21 (3) The emptiness problem is decidable for any k, provided there is only one parameterized constant. Proof: The proof of part 1 uses the undecidability of Hilbert s Tenth Problem (HTP) [16], which is to decide for a given polynomial P (x 1 ; x n ) with integer coecients whether it has a nonnegative integral root. First consider a term st(x 1 ; x n ) sx i 1 1 : x i n n of the polynomial P (x 1 ; x n ) where s = or , i 1 ; i n 0. We show how ....

Y. Matijasevic, Enumerable sets are Diophantine, Soviet Math. Dokl, 11:354{ 357, 1970.

....in general proof systems that can prove sentences about aritmethic. In particular, Theorem 32 [Gd31] There is computable function that yields for every recursive set T of axioms with APe j= T a set c such that neither T j= c nor T j= c. A much celebrated result in the eld is Theorem 33 [Mat70] It is undecidable whether a polynominal with integer coe cients has a root in N n . As an immediate consequence the 1 fragment of the theory of APe is undecidable. If we consider the structure of Peano Arithmetic without the theory is decidable: Theorem 34 (Skolem 1930 ) The theory of ....

J. V. Matijasevic. Enumerable sets are diophantine. Soviet Mathematics (Dokladi), 11(2):354357, 1970.

....terms) strong bisimulation is in Pi 0 1 , by Corollary 5.4. Now, let D consist of a two element boolean algebra, the field hZ; Theta; 0; 1i of integers and a builtin equality function [ Z 2 b; D is a computable algebra (cf. Stoltenberg Hansen and Tucker (1995) It was shown by Matijasevic (1970) that for every i 2 there exists j 2 and a polynomial U i with integer coefficients such that hx; y 1 ; y i i 2 Wn iff (9y i 1 : y j ) U i (n; x; y 1 ; y j ) 0) where W 0 ; W 1 ; is the list of all (i 2) ary recursive enumerable relations (Matijasevic and Robinson ....

Matijasevic, Y. (1970). Enumerable sets are diophantine. Soviet Math. Dokl., 11, 345--358.

....only existential questions, they do not exceed the power of passive inference devices. However, if more than one conjecture is allowed, then it is possible to learn phenomena that are not learnable by passive inductive inference devices. Using a result of Davis, Putnam, Robinson and Matijasevic [19,29], we prove that inference machines that are allowed to ask existential questions with plus and times can learn all the recursive functions. With this in mind, we concentrate on inference machines that ask questions formulated in weaker languages. If the language consists of plus and less than and ....

....using a reasonable language. Q 1 EX 0 [ EX = Q 0 EX[ j Q 1 EX 0 [ Q 1 EX 1 [ Figure 3 Summary of [ V. Asking questions using plus and times The ability to ask questions with plus and times is very powerful. The following theorem of Davis, Robinson, Putnam and Matijasevic [18,19,29] will be used to show that R, the set of recursive functions, is a member of Q 1 EX[ 2] 18 Theorem 8. 18,29] If A is an r.e. set, then there exists a polynomial p such that x 2 A , 9 z [p( z; x) 0] Moreover, given a program e, the polynomial p corresponding to the domain of e can be ....

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MATIJASEVIC, Y. Enumerable sets are diophantine. Doklady Academy Nauk. SSSR 191 (1970), 279--282. Translation in Sov. Math. Dokl. 11, 1970, pp. 354--357.

....arbitrary test and type assignment, it is in general undecidable whether the test covers the type assignment. Proof The proof is straightforward from a result, due to Matijasevic, that shows that determining the existence of (integer) solutions for arbitrary Diophantine equations is undecidable [16]. Given an arbitrary polynomial OE(x 1 ; xn ) consider the test OE(x 1 ; xn ) 6= 0. This test covers the type assignment (x 1 : integer; xn : integer) if and only if every possible assignment of integers to the variables x 1 ; xn causes the polynomial OE to take on ....

Ju. V. Matijasevic, "Enumerable Sets are Diophantine", Doklady Akademii Nauk SSSR, 191 (1970), 279-282 (in Russian; English translation in Soviet Mathematics--- Doklady, 11 (1970), 354-357).

....assignment. It turns out that in the presence of arbitrary arithmetic operations the problem is undecidable in general, even if the set of tests under consideration is a singleton (the proof is a straightforward reduction from Matijasevic s proof of the unsolvability of Hilbert s tenth problem [9]) and is co NP hard even if we restrict ourselves to finite types. We therefore have to resort to sound (but obviously incomplete) algorithms for checking coverings. A linear time algorithm for this is described in [6] Using the notion of coverings, it is straightforward to identify the ....

Ju. V. Matijasevic, "Enumerable Sets are Diophantine", Doklady Akademii Nauk SSSR, 191 (1970), 279-282 (in Russian; English translation in Soviet Mathematics---Doklady, 11 (1970), 354-357).

....stack cells. This restricted worktape is called a checking tape. Call this machine CCM. CCM acceptors without counters have been studied in [Gre68] Theorem 3. The emptiness problem for CCM acceptors is undecidable. Proof. The proof uses the undecidability of Hilbert s Tenth Problem (HTP) [Mat70], which is to decide for a given polynomial p(x 1 , x n ) with integer coe#cients whether it has a nonnegative integral root. We omit the construction, but the idea is to show that we can e#ectively construct, given a polynomial p(x 1 , x n ) a CCM acceptor M p such that p has no ....

Y. Matijasevic. Enumerable sets are Diophantine. Soviet Math. Dokl, Vol. 11, 1970, pp.354-357.

....degree n 1. The above condition represents a diophantine equation. The problem, whether there exists a procedure which in a finite number of steps enables one to determine whether or not a given diophantine equation has an integer solution is known as Hilbert s 10th problem. It has been shown ( Mat70] that this problem is undecidable. Therefore, we can express something as a condition for an FDFD, but we are not capable to provide an algorithm that allows us to decide whether this condition holds or does not hold. However, assume we define boundedint = minint, maxint and redefine the ....

J.V. Matijasevic. Enumerable Sets are Diophantine. Soviet Mathematics, Doklady, 11(2):354--358, 1970.

....functions S 1 ; S k and generalized frame sum having its parameter ranging over f1; ng . This allows a more expressive notation, generalizing the one above and leading to similar representation and decidability results, now based on the decidability result of Buchi [Buc62] 2 In [Mat70], the existence of such a predicate is proved. 8 REFERENCES ....

J.V. Matijasevic. Enumerable sets are diophantine. Soviet Math. Dokl., 11:354--358, 1970.

....examples. Hilbert s 10 th problem for the field of rationals (about algorithmic solvability of Diophantine equations over Q) inspires decision procedures for classes of equations and unsolvability results for bigger classes of first order formulas over Q, but also proofs of Matijasevic theorem [36] for various function fields over a finite field which bear relation to Q, see [42, 43] Another example is provided by the famous Riemann hypothesis. Besides partial results in its directions, analogous statements were formulated and proved for (various curves over) a finite field, see [52] The ....

Matijasevic, Y. (1970) Enumerable sets are Diophantine, Doklady AN SSSR, 191, pp.279-282.

....first order logic [10] Unification is also the basic mechanism for computing critical pairs [7] and the main inference rule of the completion procedure. On the other hand, the very old and famous 10 th Hilbert problem of solving Diophantine equations has been proved undecidable by Matijasevic [9]. The challenge now is to find the maximal subsets of Peano arithmetic for which unification is decidable. Such a candidate is actually provided by the two axioms of left and right distributivity of a symbol over a symbol . Arnborg and Tid en have shown that one sided (left or right) ....

J. V. Matijasevic. Enumerable sets are diophantine. Soviet Mathematics (Dokladi) , 11(2):354--357, 1970.

....constraints. 1 Introduction Solving Diophantine equations is a famous problem which has been widely studied starting with the Greeks. It received even more attention after Hilbert listed it as its 10th open problem. This long study resulted in 1970 in the celebrated paper of Matijasevic [15] who proved that no terminating algorithm can decide whether an arbitrary Diophantine equation has a solution or not. Research, however, proceeded actively for solving subcases, especially the subcase of linear Diophantine equations [11, 13, 2, 5, 7, 17] Indeed, solving such linear equations ....

J. V. Matijasevic. Enumerable sets are diophantine. Soviet Mathematics (Dokladi), 11(2):354--357, 1970.

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J.V. Matijasevic. Enumerable sets are diophantine. Soviet Math. Dokl., 11:354--358, 1970.

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Y. Matijasevic. "Enumerable sets are Diophantine," Soviet Math. Dokl, 11:354-357, 1970.

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Y. Matijasevic. "Enumerable sets are Diophantine," Soviet Math. Dokl, 11:354-357, 1970.

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